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Question:

Eight teams compete in a tournament. Each pair of teams plays exactly one game against each other. There are no ties. If the two possible outcomes of each game are equally likely, what is the probability that every team loses at least one game and wins at least one game?

I found this question on 2017 Fermat Competition of Waterloo University. The original solution seems a bit wordy and redundant. I wonder if there is a way to solve the probability in more simplified approaches? Original Solution (Page 8)

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    $\begingroup$ I mean, much of what they say isn't strictly necessary to say. But the solution isn't really that long. And what is it about the solution which you find "un-algebraic"? Do they not use enough symbols for your taste? $\endgroup$ – Arthur Jan 30 at 17:54
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    $\begingroup$ I can understand the solution that they provided. However, I reckon there should be algebraic solutions that are more explicit and systematic. I have found some of the official Waterloo Contest solution to be redundant before. I agree that more mathematical symbol doesn't make the solution better. But an explicit answer does. $\endgroup$ – Kevin Jan 30 at 17:59
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The probability that team${_k}$ loses its seven matches is $2^{-7}={1\over128}$. The probability that one of the eight teams loses its seven matches therefore is $8\cdot{1\over128}={1\over16}$, and similarly, the probability that one of the eight teams wins its seven matches is ${1\over16}$. The probability that at least one of these two events occurs is ${1\over16}+{1\over16}$ minus the probability that both occur at the same time. The latter is ${8\over128}\cdot{7\over64}={7\over1024}$: We can choose in $8$ ways the team${_k}$ that loses $7$ times and then in $7$ ways the team that wins against team${_k}$ and $6$ other teams.

It follows that the probability $p$ for having nothing of these is given by $$p=1-\left({1\over16}+{1\over16}-{7\over1024}\right)={903\over1024}\ .$$

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  • $\begingroup$ That's much more straightforward. Thank you so much! $\endgroup$ – Kevin Jan 30 at 20:26
  • $\begingroup$ It might be a little bit better to write $\frac 8{128} \times \frac 7{64}$ as $\mathrm{2}^{-7} \times 8$ $\times$ $\mathrm{2}^{-6} \times 7$, which would better explains the purpose of this step. $\endgroup$ – Kevin Jan 30 at 20:48
  • $\begingroup$ This is basically the exact same argument, just without all the fluff they fill the page with. $\endgroup$ – Arthur Jan 31 at 9:05

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